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Chemistry 222 Name__________________________________________ Spring 2019 80 Points Exam 2: Chapters 5.3-7 Complete problem 1 and four of problems 2-7. CLEARLY mark the problems you do not want graded. You must show your work to receive credit for problems requiring math. Report your answers with the appropriate number of significant figures. You MUST complete problem 1. (16 pts.) 1. When ammonium sulfate dissolves, both the anion and the cation can participate in acid-

base equilibria. Considering all the equilibria reflected by the equilibrium constants below, write enough equations so that you could solve for the concentration of each species in a solution that is saturated with ammonium sulfate and also contains 0.10 M potassium nitrate. You must write the charge balance expression and at least one mass balance. Identify all unknowns and write enough explicit, independent equations so that only algebra remains to solve for the unknowns. A numerical answer is not necessary. Do not consider activities.

(NH4)2SO4(s) Ksp = 276 NH4

+ Ka = 5.7 x 10-10 SO4

2- Kb = 9.8 x 10-13

H2O Kw = 1.0 x 10-14

Reactions (Unknowns are in Bold):

(NH4)2SO4(s) ⇌ 2NH4+ + SO4

2-

NH4+ ⇌ NH3 + H+

SO42- + H2O ⇌ HSO4

- + OH-

H2O ⇌ H+ + OH-

KNO3 → K+ + NO3-

So, we need a total of eight equations for our eight unknowns. Charge Balance (4) [NH4

+] + [H+] + [K+] = 2[SO42-] + [HSO4

-] + [OH-] + [NO3-]

Mass Balance: (8) total concentration of nitrogen = 2(total concentration of sulfur) [N]total = 2[S]total

[NH3] + [NH4+]= 2[SO4

2-] + 2[HSO4-]

Concentrations of Na+ and NO3 are constant [K+] = 0.10 M

[NO3-] = 0.10 M

Equilibrium constant expressions: (4)

Ka = [NH3][H+] Ksp = [NH4+]2 [SO4

2-] Kw = [H+] [OH-] [NH4

+]

Kb = [HSO4-][OH-]

[SO42-]

We now have enough equations to solve for all of the unknowns

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Do four of problems 2-7. Clearly mark the problems you do not want graded. (16 pts. ea.) 2. Consider the table of activity coefficients below. As you move from left to right across any

row on the table, the values for activity coefficient decrease. As you move down in a given column, the activity coefficient also decreases. Clearly describe the phenomena that cause these trends. Do not simply point out the trends; you must explain why the trends are observed. No calculations are necessary.

Ionic Strength Species 0.001 M 0.01 M 0.1 M

hydronium ion 0.967 0.914 0.830 nitrate ion 0.964 0.899 0.755 calcium ion 0.870 0.675 0.405 sulfate ion 0.867 0.660 0.355 phosphate ion 0.725 0.395 0.095

The cause for both trends is rooted in electrostatic interactions between the ion of interest and other materials in solution. Moving left to right across a row in the table, the ionic

strength of the solution increases. As µ increases, there is a greater probability that the ion of interest will undergo an interaction with a third-party ion in solution, thus decreasing its activity in the reaction of interest. Moving down in the table, the charge on the ion is increasing. This increase in charge provides for stronger electrostatic interactions with all other ions in solution, also reducing activity. For ions of the same charge, the ion with smaller hydrated ion size is more likely to interact with other competing ions in solution, decreasing the activity coefficient.

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3. Aziridine (C2NH5) is a monobasic weak base with a pKb of 5.96. Calculate the pH of a solution prepared by mixing 20.0 mL of 0.025 M HCl with 50.0 mL of 0.036 M aziridine and diluting the resulting solution to 100.0 mL. Do not consider autoprotolysis or activities.

Whenever you mix a strong acid with a weak base, they will react until one is completely consumed. Whatever remains after the reaction is complete will determine pH. (6) What’s left after HCl and aziridine react? (I’ll call the aziridine “B”) HCl + B → BH+ + Cl-

Start (mmol) 0.50 1.8 0 mmol 0 mmol

End (mmol) 0.50-0.50 = 0 1.8-0.5 = 1.3 0.50 0.50

Concentration(M) 0 1.3mmol/100mL = 0.013 M

0.50mmol/100 mL = 0.0050M

0.0050 M

(10) Now the equilibrium:

B + H2O BH+ + OH- Kb = [BH+][OH-] I 0.013M -- 0.0050M 0 [B] C -x -- +x +x Kb = (0.0050+x)(x) E 0.013-x -- 0.0050+x x 0.013-x

After some algebra, 0 = x2 + (0.0050+Kb)x – 0.013Kb Solving for x, we get x = [OH-] = 2.85x10-6 M, [H+] = Kw/[OH-] = 2.851x10-9 M

pOH = 5.545 pH = 14-pOH = 8.455

You can simplify the math a little if your realize this is a buffer solution and that we would not expect the equilibrium concentrations to vary much from their original concentrations, essentially assuming x<<0.0050 M. In doing so, the Kb expression becomes:

Kb = 0.0050x 0.013

Solving for x produces the same result as above since 2.85x10-6 << 0.0050 A third approach would be to recognize that, after reaction with HCl, a buffer solution is formed, and Henderson Hasslebach equation could be used:

pH = pKa + log [conjugate base] [weak acid]

pH = (14-pKb) + log [B]

[BH+]

pH = 8.04 + log [0.013] = 8.45

[0.0050]

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4. Using activities, find the mercury concentration of a solution of 0.025 M Na2SO4 saturated with Hg(SCN)2 (Ksp for Hg(SCN)2 is 2.8 x 10-20). Assume that all other salts are soluble. You may ignore the autoprotolysis of water and any acid-base character of sulfate and thiocyanate. Compare this result to that obtained if you were to ignore activities.

(4) Let's set up the system first:

Hg(SCN)2 = Hg2+ + 2 SCN2- i - 0 0 c - +x +2x e - x 2x

Ksp = AHg2+(ASCN-)2 = γHg2+[Hg2+](γSCN-[SCN-])2 = γHg2+(x)(γSCN-(2x))2= 4(γHg2+)(γSCN-)x3

x = [Hg2+] = Ksp 1/3

4(γHg2+)(γSCN-)2

(4)To find activity coefficients, we need to calculate ionic strength: [SO4

2-] = 0.025 M [Li+] = 2(0.025M) = 0.050 M

µ = 1/2([Li+] (+1)2 + [SO42-](-2)2) = 1/2(0.050(1) + 0.025(4)) = 0.075 M

(4) To find activity coefficients, we can either use the Debye-Huckel equation or interpolate

from the tabulated values, since µ = 0.075 M is not explicitly given on the table

After interpolation, for µ = 0.075 M, γHg2+ = 0.423, γSCN- = 0.785. If you use the D-H equation,

you get γHg2+ = 0.412, γSCN- = 0.783

x = [Hg2+] = Ksp 1/3 = Ksp 1/3

4(γHg2+)(γSCN-)2 4(0.423)(0.785)2

Substituting into the equation above gives [Hg2+]= 2.99 x 10-7 M (4) Ignoring activities,

[Hg2+] = x = Ksp 1/3

4 Which gives [Hg2+]= 1.91 x 10-7 M

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5. You have been given the task of explaining the similarities and differences of internal standards and standard additions to a newly hired analytical chemist, Irma Dorque. a. Describe to Irma experimental issues that may necessitate internal standards and how

they differ from those that necessitate standard additions. (5 points)

Internal standards addresses run-to-run variability and standard adds addresses matrix effects. You should elaborate on this somewhat.

b. Compare the role of the “standard” in each approach (5 points)

In internal standards the standard has a different identity than the analyte and its relative response to the analyte is important. For standard adds, the standard has the same identity of the analyte and is used to evaluate the sensitivity of the measurement for the analyte.

c. For either internal standards or standard additions, describe how an experiment would be run in practice and how an unknown analyte concentration could be determined from the results obtained. (hint: you will make more than two measurements) (6 points)

You should discuss solution and sample preparation. Generation of a calibration curve. How the calibration response allows for the determination of the analyte response and how the analyte concentration is determined from the analyte response. In practice the two-data point approach would not be used.

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6. Complete both parts a and b. (8 points each) a. A saturated solution of Ag2CO3 (Ksp = 8.1 x 10-12) that originally had a volume of 1.00 L is

allowed to evaporate until the solution volume is 0.500 L. How does the new concentration of Ag+ compare to the concentration in the original solution? Clearly justify your response. Do not consider activities.

Since the solution was initially saturated, a decrease in solution volume will force an increase in the concentrations of Ba2+ and SO4

2- in solution, leading to supersaturation. This increase will cause the Ksp for barium sulfate to be exceeded, resulting in the precipitation of BaSO4, and a decrease in the concentration of Ba2+ and SO4

2- back to their original values. Therefore, the [Ba2+] after evaporation will be the same as its initial concentration.

b. In determining the pH of a solution that contains and 0.10 M formic acid (Ka = 1.8 x 10-4) and 0.10 M HF (Ka = 6.8 x 10-4), why is the ICE table approach not a valid strategy to employ? What approach should be used instead? Do not consider activities.

Since we have two weak acids in solution, we must account for them simultaneously. That is, since H+ appears in both equilibria and since there can only be one H+ concentration in solution, that single concentration must satisfy both equilibrium constant expressions simultaneously. The ICE table approach only addresses one equilibrium at a time. Therefore, if you solve the ICE table for HF to determine [H+], the concentration you arrive at will not correspond to the equilibrium concentration for formic acid. You should use the systematic approaching using mass and charge balance instead.

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7. Is it possible to perform a 99.99 % complete separation of barium and silver by precipitation with carbonate if both Ba2+ and Ag+ are present initially at 0.020 M? Justify your decision. Do not consider activities.

BaCO3 Ksp = 5.0 x 10-9

Ag2CO3 Ksp = 8.1 x 10-12

Two equilibria to consider:

BaCO3 = Ca2+ + CO32- Ksp = 5.0 x 10-9

Ag2CO3 = 2Ag+ + CO32- Ksp = 8.1 x 10-12

1. What [CO3

2-] is needed to lower each ion's concentration to 0.01% of its initial value? Target [Ba2+] = 0.0001(0.020M) = 2.0 x 10-6 M, Target [Ag+] = 2.0 x 10-6 M

Ba2+: [CO32-] = Ksp = 5.0 x 10-9 = 0.0025 M CO3

2-

[Ba2+] 2.0 x 10-6 M

Ag+: [CO32-] = Ksp = 8.1 x 10-12 = 2.025 M CO3

2-

[Ag+]2 (2.0 x 10-6 M)2 So, Ba2+ will precipitate first. 2. Will Ag+ precipitate if [CO3

2-] = 0.0025 M? Q = [Ag+]2[CO3

2-] = (0.020 M)2(0.0025 M) = 1.0 x 10-6 Since Q>Ksp for Ag2 CO3, Ag+ will precipitate before [Ba2+] = 2.0 x 10-6 M. Separation is not feasible.

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Possibly Useful Information

KaKb = KW = 1.0 x 10-14 pH = -log [H+]

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